Question

A sample of an ideal gas undergoes an isothermal process as shown by the curve $AB$ in the $pV$ diagram. If $\Delta Q$ , $\Delta U$ and $\Delta W$ represent the amount of heat absorbed the change in internal energy and the work done respectively, then which of the following statement is correct?

• A. $\Delta Q=+ve,\Delta U=0,\Delta W=-ve$
• B. $\Delta Q=+ve,\Delta U=0,\Delta W=+ve$
• C. $\Delta Q=+ve,\Delta U=0,\Delta W=0$
• D. $\Delta Q=+ve,\Delta U=+ve,\Delta W=+ve$

Answer 1

Answer: (B)

From the graph of an ideal gas given in the question, the process is an isothermal expansion because volume of the ideal gas increases
Given, $V_B>V_A$ and $p_B<P_A$
$\because \Delta T=0$ (isothermal process)
$\therefore$ Change in internal energy
$\Delta U=0\left(\because \Delta U\propto \Delta T\right)$
Since we know, in isothermal process
$\Delta Q=\Delta W=n\cdot RT\ln \left(\frac{V_f}{V_p}\right)$
or $\Delta Q=\Delta W=nRT\ln \left(\frac{V_B}{V_A}\right)$
Thus, $\Delta Q=+ve$ and $\Delta w=+ve$
because $V_B>V_A$ (from the graph)